Maharashtra State Board course SSC (English Medium) Class 10th Board Exam
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Prave That: Sec θ + Tan θ = Cos θ 1 − Sin θ - Geometry

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Question

Prove that:

\[\sec\theta + \tan\theta = \frac{\cos\theta}{1 - \sin\theta}\]

Solution

\[\sec\theta + \tan\theta\]

\[ = \frac{1}{\cos\theta} + \frac{\sin\theta}{\cos\theta}\]

\[ = \frac{1 + \sin\theta}{\cos\theta}\]

\[ = \frac{\cos\theta\left( 1 + \sin\theta \right)}{\cos^2 \theta}\]

\[ = \frac{\cos\theta\left( 1 + \sin\theta \right)}{1 - \sin^2 \theta} \left( \sin^2 \theta + \cos^2 \theta = 1 \right)\]

\[= \frac{\cos\theta\left( 1 + \sin\theta \right)}{\left( 1 + \sin\theta \right)\left( 1 - \sin\theta \right)}\]
\[ = \frac{\cos\theta}{1 - \sin\theta}\]

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APPEARS IN

 Balbharati Solution for Balbharati Class 10 Mathematics 2 Geometry (2018 to Current)
Chapter 6: Trigonometry
Practice set 6.1 | Q: 6.08 | Page no. 131
Solution Prave That: Sec θ + Tan θ = Cos θ 1 − Sin θ Concept: Application of Trigonometry.
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